What is the integral from 0 to t of (sin(u))/u ?

The integral from 0 to t of (sin(u))/u is \Si(t)

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integral from 0 to t of (sin(u))/u

\int_{0}^{t} \frac{sin ( u )}{u} d u

Si (t)

Solution steps

\int_{0}^{t} \frac{sin ( u )}{u} d u

Use the nonelementary integral:

\int \frac{sin ( x )}{x} d x = Si (x)

= [Si (u)]_{0}^{t}

Compute the boundaries:

Si (t)

= Si (t)

\int_{\frac{π}{2}}^{π} x d x

\int_{- 5}^{5} (5 x - 2) d x

\int_{2}^{5} 4^{5 x} d x

\int_{0}^{2} (3 x + 1) d x

\int_{8}^{9} 1 d t

What is the integral from 0 to t of (sin(u))/u ?
The integral from 0 to t of (sin(u))/u is \Si(t)